hdu 3530 Subsequence(单调队列)

来源：互联网发布：浙江教师网络培训平台编辑：程序博客网时间：2024/05/21 22:35

题意：求区间最值差在[m,k]范围内的最长区间。

解法类似于hdu 4123 Bob’s Race(单调队列或者rmq)。稍微有一点点的不一样而已。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std ;const int maxn = 111111 ;int n , m , num[maxn] , k ;struct Deque {int val[2][maxn], pos[2][maxn];int star[2] , tail[2] ;void init () {star[0] = 1 ;tail[0] = 0 ;star[1] = 1 ;tail[1] = 0 ;}int judge (int c, int v) {return c ^ (val[c][tail[c]] < v);}void push (int id, int v) {int i ;//printf ( "id = %d , v = %d\n" , id , v ) ;for ( i = 0 ; i < 2 ; i ++ ) {while ( star[i] <= tail[i] && judge ( i , v ) )tail[i] -- ;val[i][++tail[i]] = v ;pos[i][tail[i]] = id ;}}void pop ( int c ) {if ( star[c] <= tail[c] ) star[c] ++ ;}int gao ( int &last ) {while ( (val[0][star[0]] - val[1][star[1]] > k ) ) {if ( pos[1][star[1]] < pos[0][star[0]] ) {last = pos[1][star[1]] ;pop ( 1 ) ;}else {last = pos[0][star[0]] ;pop ( 0 ) ;}}if ( val[0][star[0]] - val[1][star[1]] < m ) return -1 ;return last ;}} q ;int main () {int i , j ;while ( scanf ( "%d%d%d" , &n , &m , &k ) != EOF ) {    for ( i = 1 ; i <= n ; i ++ )            scanf ( "%d" , &num[i] ) ;        q.init () ;        int ans = 0 , last = 0 ;        for ( i = 1 ; i <= n ; i ++ ) {            q.push ( i , num[i] ) ;            int k = q.gao ( last ) ;            if ( k == -1 ) continue ;            last = k ;            ans = max ( ans , i - last ) ;        }        printf ( "%d\n" , ans ) ;}return 0 ;}